3.189 \(\int \frac {1}{x^4 (a+b x^2)^3} \, dx\)

Optimal. Leaf size=87 \[ \frac {35 b^{3/2} \tan ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{8 a^{9/2}}+\frac {35 b}{8 a^4 x}-\frac {35}{24 a^3 x^3}+\frac {7}{8 a^2 x^3 \left (a+b x^2\right )}+\frac {1}{4 a x^3 \left (a+b x^2\right )^2} \]

[Out]

-35/24/a^3/x^3+35/8*b/a^4/x+1/4/a/x^3/(b*x^2+a)^2+7/8/a^2/x^3/(b*x^2+a)+35/8*b^(3/2)*arctan(x*b^(1/2)/a^(1/2))
/a^(9/2)

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Rubi [A]  time = 0.03, antiderivative size = 87, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 3, integrand size = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.231, Rules used = {290, 325, 205} \[ \frac {35 b^{3/2} \tan ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{8 a^{9/2}}+\frac {7}{8 a^2 x^3 \left (a+b x^2\right )}+\frac {35 b}{8 a^4 x}-\frac {35}{24 a^3 x^3}+\frac {1}{4 a x^3 \left (a+b x^2\right )^2} \]

Antiderivative was successfully verified.

[In]

Int[1/(x^4*(a + b*x^2)^3),x]

[Out]

-35/(24*a^3*x^3) + (35*b)/(8*a^4*x) + 1/(4*a*x^3*(a + b*x^2)^2) + 7/(8*a^2*x^3*(a + b*x^2)) + (35*b^(3/2)*ArcT
an[(Sqrt[b]*x)/Sqrt[a]])/(8*a^(9/2))

Rule 205

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]*ArcTan[x/Rt[a/b, 2]])/a, x] /; FreeQ[{a, b}, x]
&& PosQ[a/b]

Rule 290

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> -Simp[((c*x)^(m + 1)*(a + b*x^n)^(p + 1))/(
a*c*n*(p + 1)), x] + Dist[(m + n*(p + 1) + 1)/(a*n*(p + 1)), Int[(c*x)^m*(a + b*x^n)^(p + 1), x], x] /; FreeQ[
{a, b, c, m}, x] && IGtQ[n, 0] && LtQ[p, -1] && IntBinomialQ[a, b, c, n, m, p, x]

Rule 325

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[((c*x)^(m + 1)*(a + b*x^n)^(p + 1))/(a*
c*(m + 1)), x] - Dist[(b*(m + n*(p + 1) + 1))/(a*c^n*(m + 1)), Int[(c*x)^(m + n)*(a + b*x^n)^p, x], x] /; Free
Q[{a, b, c, p}, x] && IGtQ[n, 0] && LtQ[m, -1] && IntBinomialQ[a, b, c, n, m, p, x]

Rubi steps

\begin {align*} \int \frac {1}{x^4 \left (a+b x^2\right )^3} \, dx &=\frac {1}{4 a x^3 \left (a+b x^2\right )^2}+\frac {7 \int \frac {1}{x^4 \left (a+b x^2\right )^2} \, dx}{4 a}\\ &=\frac {1}{4 a x^3 \left (a+b x^2\right )^2}+\frac {7}{8 a^2 x^3 \left (a+b x^2\right )}+\frac {35 \int \frac {1}{x^4 \left (a+b x^2\right )} \, dx}{8 a^2}\\ &=-\frac {35}{24 a^3 x^3}+\frac {1}{4 a x^3 \left (a+b x^2\right )^2}+\frac {7}{8 a^2 x^3 \left (a+b x^2\right )}-\frac {(35 b) \int \frac {1}{x^2 \left (a+b x^2\right )} \, dx}{8 a^3}\\ &=-\frac {35}{24 a^3 x^3}+\frac {35 b}{8 a^4 x}+\frac {1}{4 a x^3 \left (a+b x^2\right )^2}+\frac {7}{8 a^2 x^3 \left (a+b x^2\right )}+\frac {\left (35 b^2\right ) \int \frac {1}{a+b x^2} \, dx}{8 a^4}\\ &=-\frac {35}{24 a^3 x^3}+\frac {35 b}{8 a^4 x}+\frac {1}{4 a x^3 \left (a+b x^2\right )^2}+\frac {7}{8 a^2 x^3 \left (a+b x^2\right )}+\frac {35 b^{3/2} \tan ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{8 a^{9/2}}\\ \end {align*}

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Mathematica [A]  time = 0.05, size = 79, normalized size = 0.91 \[ \frac {35 b^{3/2} \tan ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{8 a^{9/2}}+\frac {-8 a^3+56 a^2 b x^2+175 a b^2 x^4+105 b^3 x^6}{24 a^4 x^3 \left (a+b x^2\right )^2} \]

Antiderivative was successfully verified.

[In]

Integrate[1/(x^4*(a + b*x^2)^3),x]

[Out]

(-8*a^3 + 56*a^2*b*x^2 + 175*a*b^2*x^4 + 105*b^3*x^6)/(24*a^4*x^3*(a + b*x^2)^2) + (35*b^(3/2)*ArcTan[(Sqrt[b]
*x)/Sqrt[a]])/(8*a^(9/2))

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fricas [A]  time = 1.01, size = 238, normalized size = 2.74 \[ \left [\frac {210 \, b^{3} x^{6} + 350 \, a b^{2} x^{4} + 112 \, a^{2} b x^{2} - 16 \, a^{3} + 105 \, {\left (b^{3} x^{7} + 2 \, a b^{2} x^{5} + a^{2} b x^{3}\right )} \sqrt {-\frac {b}{a}} \log \left (\frac {b x^{2} + 2 \, a x \sqrt {-\frac {b}{a}} - a}{b x^{2} + a}\right )}{48 \, {\left (a^{4} b^{2} x^{7} + 2 \, a^{5} b x^{5} + a^{6} x^{3}\right )}}, \frac {105 \, b^{3} x^{6} + 175 \, a b^{2} x^{4} + 56 \, a^{2} b x^{2} - 8 \, a^{3} + 105 \, {\left (b^{3} x^{7} + 2 \, a b^{2} x^{5} + a^{2} b x^{3}\right )} \sqrt {\frac {b}{a}} \arctan \left (x \sqrt {\frac {b}{a}}\right )}{24 \, {\left (a^{4} b^{2} x^{7} + 2 \, a^{5} b x^{5} + a^{6} x^{3}\right )}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^4/(b*x^2+a)^3,x, algorithm="fricas")

[Out]

[1/48*(210*b^3*x^6 + 350*a*b^2*x^4 + 112*a^2*b*x^2 - 16*a^3 + 105*(b^3*x^7 + 2*a*b^2*x^5 + a^2*b*x^3)*sqrt(-b/
a)*log((b*x^2 + 2*a*x*sqrt(-b/a) - a)/(b*x^2 + a)))/(a^4*b^2*x^7 + 2*a^5*b*x^5 + a^6*x^3), 1/24*(105*b^3*x^6 +
175*a*b^2*x^4 + 56*a^2*b*x^2 - 8*a^3 + 105*(b^3*x^7 + 2*a*b^2*x^5 + a^2*b*x^3)*sqrt(b/a)*arctan(x*sqrt(b/a)))
/(a^4*b^2*x^7 + 2*a^5*b*x^5 + a^6*x^3)]

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giac [A]  time = 0.64, size = 71, normalized size = 0.82 \[ \frac {35 \, b^{2} \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{8 \, \sqrt {a b} a^{4}} + \frac {11 \, b^{3} x^{3} + 13 \, a b^{2} x}{8 \, {\left (b x^{2} + a\right )}^{2} a^{4}} + \frac {9 \, b x^{2} - a}{3 \, a^{4} x^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^4/(b*x^2+a)^3,x, algorithm="giac")

[Out]

35/8*b^2*arctan(b*x/sqrt(a*b))/(sqrt(a*b)*a^4) + 1/8*(11*b^3*x^3 + 13*a*b^2*x)/((b*x^2 + a)^2*a^4) + 1/3*(9*b*
x^2 - a)/(a^4*x^3)

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maple [A]  time = 0.02, size = 79, normalized size = 0.91 \[ \frac {11 b^{3} x^{3}}{8 \left (b \,x^{2}+a \right )^{2} a^{4}}+\frac {13 b^{2} x}{8 \left (b \,x^{2}+a \right )^{2} a^{3}}+\frac {35 b^{2} \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{8 \sqrt {a b}\, a^{4}}+\frac {3 b}{a^{4} x}-\frac {1}{3 a^{3} x^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/x^4/(b*x^2+a)^3,x)

[Out]

-1/3/a^3/x^3+3*b/a^4/x+11/8/a^4*b^3/(b*x^2+a)^2*x^3+13/8/a^3*b^2/(b*x^2+a)^2*x+35/8/a^4*b^2/(a*b)^(1/2)*arctan
(1/(a*b)^(1/2)*b*x)

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maxima [A]  time = 2.96, size = 86, normalized size = 0.99 \[ \frac {105 \, b^{3} x^{6} + 175 \, a b^{2} x^{4} + 56 \, a^{2} b x^{2} - 8 \, a^{3}}{24 \, {\left (a^{4} b^{2} x^{7} + 2 \, a^{5} b x^{5} + a^{6} x^{3}\right )}} + \frac {35 \, b^{2} \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{8 \, \sqrt {a b} a^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^4/(b*x^2+a)^3,x, algorithm="maxima")

[Out]

1/24*(105*b^3*x^6 + 175*a*b^2*x^4 + 56*a^2*b*x^2 - 8*a^3)/(a^4*b^2*x^7 + 2*a^5*b*x^5 + a^6*x^3) + 35/8*b^2*arc
tan(b*x/sqrt(a*b))/(sqrt(a*b)*a^4)

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mupad [B]  time = 4.67, size = 80, normalized size = 0.92 \[ \frac {\frac {7\,b\,x^2}{3\,a^2}-\frac {1}{3\,a}+\frac {175\,b^2\,x^4}{24\,a^3}+\frac {35\,b^3\,x^6}{8\,a^4}}{a^2\,x^3+2\,a\,b\,x^5+b^2\,x^7}+\frac {35\,b^{3/2}\,\mathrm {atan}\left (\frac {\sqrt {b}\,x}{\sqrt {a}}\right )}{8\,a^{9/2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(x^4*(a + b*x^2)^3),x)

[Out]

((7*b*x^2)/(3*a^2) - 1/(3*a) + (175*b^2*x^4)/(24*a^3) + (35*b^3*x^6)/(8*a^4))/(a^2*x^3 + b^2*x^7 + 2*a*b*x^5)
+ (35*b^(3/2)*atan((b^(1/2)*x)/a^(1/2)))/(8*a^(9/2))

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sympy [A]  time = 0.48, size = 138, normalized size = 1.59 \[ - \frac {35 \sqrt {- \frac {b^{3}}{a^{9}}} \log {\left (- \frac {a^{5} \sqrt {- \frac {b^{3}}{a^{9}}}}{b^{2}} + x \right )}}{16} + \frac {35 \sqrt {- \frac {b^{3}}{a^{9}}} \log {\left (\frac {a^{5} \sqrt {- \frac {b^{3}}{a^{9}}}}{b^{2}} + x \right )}}{16} + \frac {- 8 a^{3} + 56 a^{2} b x^{2} + 175 a b^{2} x^{4} + 105 b^{3} x^{6}}{24 a^{6} x^{3} + 48 a^{5} b x^{5} + 24 a^{4} b^{2} x^{7}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x**4/(b*x**2+a)**3,x)

[Out]

-35*sqrt(-b**3/a**9)*log(-a**5*sqrt(-b**3/a**9)/b**2 + x)/16 + 35*sqrt(-b**3/a**9)*log(a**5*sqrt(-b**3/a**9)/b
**2 + x)/16 + (-8*a**3 + 56*a**2*b*x**2 + 175*a*b**2*x**4 + 105*b**3*x**6)/(24*a**6*x**3 + 48*a**5*b*x**5 + 24
*a**4*b**2*x**7)

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